The difference between the maximum value and minimum value of the objective function $z = 3x + 5y$ subject to the constraints $x + 3y \leqslant 60$,$x + y \geqslant 10$,$x - y = 0$,and $x, y \geqslant 0$ is:

The maximum value of $Z=5x+4y$,subject to the constraints $y \leq 2x$,$x \leq 2y$,$x+y \leq 3$,$x \geq 0$,$y \geq 0$ is:

The minimum value of $z = 3x + 5y$, subject to constraints $x \leq 80$, $y \geq 60$, $x + y \leq 200$ and $x, y \geq 0$ occurs at the point:

$A$ factory manufactures two types of screws, $A$ and $B$. Each type of screw requires the use of two machines, an automatic and a hand-operated one. It takes $4 \, \text{minutes}$ on the automatic and $6 \, \text{minutes}$ on the hand-operated machine to manufacture a package of screws $A$, while it takes $6 \, \text{minutes}$ on the automatic and $3 \, \text{minutes}$ on the hand-operated machine to manufacture a package of screws $B$. Each machine is available for at most $4 \, \text{hours}$ on any day. The manufacturer can sell a package of screws $A$ at a profit of $Rs. \, 7$ and screws $B$ at a profit of $Rs. \, 10$. Assuming that he can sell all the screws he manufactures, how many packages of each type should the factory owner produce in a day in order to maximize his profit? Determine the maximum profit.

$A$ manufacturer makes two types of toys $A$ and $B$. Three machines are needed for this purpose and the time (in $minutes$) required for each toy on the machines is given below:

Types of Toys	Machine-$I$	Machine-$II$	Machine-$III$
$A$	$12$	$18$	$6$
$B$	$6$	$0$	$9$

Each machine is available for a maximum of $6 \, hours$ $(360 \, minutes)$ per day. If the profit on each toy of type $A$ is $Rs. \, 7.50$ and that on each toy of type $B$ is $Rs. \, 5$,find the number of toys of each type that should be manufactured in a day to get maximum profit.

For the linear programming problem, $x + 2y \leq 10$, $3x + y \leq 12$, $x, y \geq 0$, the maximum value of $z = 5x + 10y$ occurs at every point on the line segment joining the points..